A 1-pound bag of peanuts contained 430 peanuts. The distribution of weights (in grams) of the peanuts isgiven below . What is the sample standard deviation of the weight of a peanut? $$\begin{array} { | l | l | }
\hline \begin{array} { l }
\text { Weights } \
\text { (grams) }
\end{array} & { \text { Frequency } } \
\hline 0.755 - 0.814 & 3 \
\hline 0.815 - 0.874 & 2 \
\hline 0.875 - 0.934 & 2 \
\hline 0.935 - 0.994 & 2 \
\hline 0.995 - 1.054 & 168 \
\hline 1.055 - 1.114 & 241 \
\hline 1.115 - 1.174 & 12 \
\hline
\end{array}$$
A) 0.045 g
B) 0.002 g
C) 0.209 g
D) 0.000004 g

Question

A 1-pound bag of peanuts contained 430 peanuts. The distribution of weights (in grams) of the peanuts isgiven below . What is the sample standard deviation of the weight of a peanut? $$\begin{array} { | l | l | } 
\hline \begin{array} { l } 
\text { Weights } \
\text { (grams) }
\end{array} &  { \text { Frequency } } \
\hline 0.755 - 0.814 & 3 \
\hline 0.815 - 0.874 & 2 \
\hline 0.875 - 0.934 & 2 \
\hline 0.935 - 0.994 & 2 \
\hline 0.995 - 1.054 & 168 \
\hline 1.055 - 1.114 & 241 \
\hline 1.115 - 1.174 & 12 \
\hline
\end{array}$$
A) 0.045 g
B) 0.002 g
C) 0.209 g
D) 0.000004 g

Quizplus · Accepted Answer

The sample standard deviation is 0.045 g. This can be calculated using the formula:s = sqrt(sum((x - mean)^2 / (n - 1)))where x is the weight of a peanut, mean is the mean weight of a peanut, and n is the number of peanuts.The mean weight of a peanut is:mean = (0.755 + 0.815 + 0.875 + 0.935 + 0.995 + 1.055 + 1.115) / 7 = 0.995 gThe sample standard deviation is:s = sqrt((sum((x - 0.995)^2 / (430 - 1)))) = 0.045 g

A 1-Pound Bag of Peanuts Contained 430 Peanuts A) 0045 G B) 0

A 1-Pound Bag of Peanuts Contained 430 Peanuts A) 0045 G
B) 0